How much ethyl alcohol $\left({\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\right)$ must be added to 1 L of water so that solution will not freeze at $-6^{\mathrm{o}}\mathrm{F}$?

$({\mathrm{K}}_{\mathrm{f}}\text{ of }{\mathrm{H}}_2\mathrm{O}=1{.86}^{\mathrm{o}}{\text{mol}}^{-1}\text{kg})$

$-6^{o}F$ is to be converted into ${ }^{o}C$

$\begin{array}{l}\frac{\mathrm{F}-32}{180}= \frac{\mathrm{C}}{100}\\ \frac{-4-32}{180}=\frac{\mathrm{C}}{100}\\ \therefore \mathrm{C}=-{20}^{\mathrm{o}}\mathrm{C}\\ \therefore {\mathrm{\Delta T}}_{\mathrm{f}}={20}^{\mathrm{o}}\mathrm{C}\end{array}$

$\begin{array}{l}1 \mathrm{L} {\mathrm{H}}_2\mathrm{O}=1000 \mathrm{g} \mathrm{water}\\ {\mathrm{\Delta T}}_{\mathrm{f}}=\frac{1000\mathrm{\hspace{0.17em}}{\mathrm{K}}_{\mathrm{f}}{\mathrm{w}}_1}{{\mathrm{m}}_1{\mathrm{w}}_2}\\ {\mathrm{w}}_1\mathrm{\hspace{0.17em}}(\mathrm{ethyl} \mathrm{alcohol})=\frac{{\mathrm{\Delta T}}_{\mathrm{f}}{\mathrm{m}}_1{\mathrm{w}}_2}{1000\mathrm{\hspace{0.17em}}{\mathrm{K}}_{\mathrm{f}}}=\frac{20\times 46\times 1000}{1000\times 1.86}\\ =\frac{920000}{1860}=494.6g\end{array}$